A110037 Signed version of A090678 and congruent to A088567 mod 2.

1, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 1, 0, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 1, 0, -1, 0, 0, 1, 0, -1, 0

Offset: 0

Author

Paul D. Hanna, Jul 09 2005

Keywords

• sign
• changed

Comments

a(n) = (-1)^[n/2]*A090678(n) = A088567(n) mod 2, where A088567(n) equals the number of "non-squashing" partitions of n. a(n) = -A110036(n)/2 for n>=2, where the A110036 gives the partial quotients of the continued fraction expansion of 1 + Sum_{n>=0} 1/x^(2^n).

Crossrefs

Cf. A110036, A090678, A088567, A073089.

Programs

PARI

{a(n)=polcoeff(A=1+x-x^2*(1+x)/(1+x^2)+ sum(k=1,#binary(n),x^(3*2^(k-1))/prod(j=0,k,1+x^(2^j)+x*O(x^n))),n)}

Formula

G.f.: A(x) = 1+x - x^2*(1+x)/(1+x^2) + Sum_{k>=1} x^(3*2^(k-1))/Product_{j=0..k} (1+x^(2^j)).

Conjecture: a(n) = A073089(n) - A073089(n+1) for n >= 2. - Alan Michael Gómez Calderón, Aug 19 2025